X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Received: from imr-da05.mx.aol.com ([205.188.105.147] verified) by logan.com (CommuniGate Pro SMTP 5.3.9) with ESMTP id 4451439 for flyrotary@lancaironline.net; Wed, 01 Sep 2010 00:22:25 -0400 Received-SPF: pass receiver=logan.com; client-ip=205.188.105.147; envelope-from=WRJJRS@aol.com Received: from imo-da04.mx.aol.com (imo-da04.mx.aol.com [205.188.169.202]) by imr-da05.mx.aol.com (8.14.1/8.14.1) with ESMTP id o814LjIA030427 for ; Wed, 1 Sep 2010 00:21:45 -0400 Received: from WRJJRS@aol.com by imo-da04.mx.aol.com (mail_out_v42.9.) id q.d18.58f7c434 (45494) for ; Wed, 1 Sep 2010 00:21:39 -0400 (EDT) Received: from magic-d24.mail.aol.com (magic-d24.mail.aol.com [172.19.146.158]) by cia-mc08.mx.aol.com (v129.4) with ESMTP id MAILCIAMC081-b1b64c7dd4d3249; Wed, 01 Sep 2010 00:21:39 -0400 From: WRJJRS@aol.com Message-ID: <11e90d.35b4db9a.39af2ed3@aol.com> Date: Wed, 1 Sep 2010 00:21:39 EDT Subject: Re: [FlyRotary] Radiator Math To: flyrotary@lancaironline.net MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="part1_11e90d.35b4db9a.39af2ed3_boundary" X-Mailer: 9.0 SE for Windows sub 5046 X-AOL-IP: 75.208.218.72 X-Spam-Flag:NO X-AOL-SENDER: WRJJRS@aol.com --part1_11e90d.35b4db9a.39af2ed3_boundary Content-Type: text/plain; charset="UTF-8" Content-Transfer-Encoding: quoted-printable Content-Language: en =20 In a message dated 8/31/2010 7:40:36 PM Pacific Daylight Time, =20 candtmallory@embarqmail.com writes: =20 Let me start with my old radiator. The core is 8 x 15 x 5. It is a=20 double pass radiator. My inlet is 36 sq in. I think the radiator is jus= t too=20 thick for good enough air flow at the speeds we are climbing and cruising= ,=20 not to mention ground ops.=20 So I geeked out, and did the math that I have found to figure out the=20 correct size. Those who have done this before, please check my math.=20 I found the following requirements/suggestions during my research:=20 1.2 sq in of face per cubic in of displacement=20 1.2*3*39.9 =3D 143.64 sq in face=20 2.1 cubic in per HP=20 2.1*210 =3D 441 cubic in=20 2.48 cubic in per HP=20 2.48*210=3D521 cubic in=20 Inlet should be 15% of face=20 36 sq in inlet =3D 240 sq in face=20 Since my oil cooler is on the side, and works fine, I get to use the=20 entire bottom of the engine for the radiator. This gives me a max space= of 16 x=20 18 for 288 sq in face. Using a 2.5 in thick core 288*2.5 =3D 720 cubic= in=20 of radiator. This should do the trick. I could even make it a little=20 smaller to ensure easy clearance. My inlet may be a little smaller than= the=20 radiator can handle, but I don=E2=80=99t see how it can hurt to have a sl= ightly larger=20 radiator than the inlet can handle.=20 Along the way I found a reference that said the heat from 13.7 HP is shed= =20 for every 1*C temp differential for every sq ft of intake. Assume=20 200F(93C) coolant and a hot day, 90F(30C) I get:=20 (13.7/144) inlet *63 =3D 210 so inlet =3D 35 sq in=20 So up to now I=E2=80=99m feeling pretty good about the math, but please= let me=20 know if I messed it up.=20 I will have to use the wedge shaped duct to move the air through the=20 radiator. So to figure out the height of the duct from the radiator, I= again=20 used 15% of the facial area, then divided by the width.=20 16 x 18 x 15% =3D 43.2=20 43.2 / 16 =3D 2.7=20 So Do I really make the front of the wedge only 2.7 inches tall? This=20 seems pretty small. It would result in a really long, thin triangle. I= have=20 a max of 6.5 inches available, so can easily make it bigger.=20 Thanks for the help.=20 Chris Chris, If I recall correctly the wedge duct requires a bit more inlet than that= =20 about 20% is minimum. More important than the inlet, the area behind the= =20 radiator is critical. You need 4 inches of unobstructed area if using a = wedge=20 duct and more is better. Bill Jepson --part1_11e90d.35b4db9a.39af2ed3_boundary Content-Type: text/html; charset="UTF-8" Content-Transfer-Encoding: quoted-printable Content-Language: en <= FONT id=3Drole_document face=3DArial color=3D#000000 size=3D2>
In a message dated 8/31/2010 7:40:36 PM Pacific Daylight Time,=20 candtmallory@embarqmail.com writes:

Let me start with my old radiator.  The core= is 8 x 15=20 x 5.  It is a double pass radiator.  My inlet is 36 sq in.&nbs= p; I=20 think the radiator is just too thick for good enough air flow at the spe= eds we=20 are climbing and cruising, not to mention ground ops.

 

So I geeked out, and did the math that I have found= to=20 figure out the correct size.  Those who have done this before, plea= se=20 check my math.

 

I found the following requirements/suggestions duri= ng my=20 research:

 

1.2 sq in of face per cubic in of=20 displacement

        = ;       =20 1.2*3*39.9 =3D 143.64 sq in face

2.1 cubic in per HP

        = ;       =20 2.1*210 =3D 441 cubic in

2.48 cubic in per HP

        = ;       =20 2.48*210=3D521 cubic in

Inlet should be 15% of face

        = ;       =20 36 sq in inlet =3D 240 sq in face

Since my oil cooler is on the side, and works fine,= I get=20 to use the entire bottom of the engine for the radiator.  This give= s me a=20 max space of 16 x 18 for 288 sq in face.  Using a 2.5 in thick core= =20 288*2.5 =3D 720 cubic in of radiator.  This should do the trick.&nb= sp; I=20 could even make it a little smaller to ensure easy clearance.  My= inlet=20 may be a little smaller than the radiator can handle, but I don=E2=80=99= t see how it=20 can hurt to have a slightly larger radiator than the inlet can=20 handle.

 

Along the way I found a reference that said the hea= t from=20 13.7 HP is shed for every 1*C temp differential for every sq ft of=20 intake.  Assume 200F(93C) coolant and a hot day, 90F(30C) I=20 get:

        = ;       =20 (13.7/144) inlet *63 =3D 210  so inlet =3D 35 sq in

So up to now I=E2=80=99m feeling pretty good about= the math, but=20 please let me know if I messed it up.

 

I will have to use the wedge shaped duct to move th= e air=20 through the radiator.  So to figure out the height of the duct from= the=20 radiator, I again used 15% of the facial area, then divided by the=20 width.

        = ;       =20 16 x 18 x 15% =3D 43.2

        = ;       =20 43.2 / 16 =3D 2.7

So Do I really make the front of the wedge only 2.7= inches=20 tall?  This seems pretty small.  It would result in a really= long,=20 thin triangle.  I have a max of 6.5 inches available, so can easily= make=20 it bigger.

 

Thanks for the help.

 

Chris

Chris,
If I recall correctly the wedge duct requires a bit more inlet than= that=20 about 20% is minimum. More important than the inlet, the area behind the= =20 radiator is critical. You need 4 inches of unobstructed area if usi= ng a=20 wedge duct and more is better.
Bill Jepson
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